What is the integral of 1/((2-x)) ?

The integral of 1/((2-x)) is -ln|2-x|+C

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integral of 1/((2-x))

\int \frac{1}{( 2 - x )} d x

- ln ∣ 2 - x ∣ + C

Solution steps

\int \frac{1}{2 - x} d x

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 2 - x

= - ln ∣ 2 - x ∣

Add a constant to the solution

= - ln ∣ 2 - x ∣ + C

\int \frac{1}{3 + x - x ^{2}} d x

(x^{3} - 2 x + 1) (3 x^{3} + 2 x^{2} - 5 x)

y = \frac{( x + 1 ) ^{2}}{x + 1}

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} d y + \frac{t}{2 y ^{4}} d t = 0

n = 0 \sum \infty (\frac{3}{5})^{n}